A277365
a(n) is the smallest number k such that f(k) = f(n) + f(n+1) and g(k) = g(n) + g(n+1), where f(n) (resp. g(n)) is the number of halving (resp. tripling) steps to reach 1 in the Collatz ('3x+1') problem.
Original entry on oeis.org
2, 6, 12, 20, 34, 49, 56, 72, 98, 112, 144, 176, 196, 228, 224, 272, 344, 406, 392, 384, 448, 520, 576, 688, 688, 772, 913, 912, 912, 1028, 992, 1040, 1220, 1152, 1376, 1624, 1624, 1708, 1624, 1728, 1728, 1824, 2160, 2080, 2080, 2215, 2559, 2752, 2884, 2884, 2752
Offset: 1
a(3) = 12 because (A006666(3), A006667(3)) = (f(3), g(3)) = (5, 2) => f(12) = f(3) + f(4) = 5 + 2 = 7 and g(12) = g(3) + g(4) = 2 + 0 = 2.
-
nn:=10^6:U:=array(1..nn):V:=array(1..nn):
for i from 1 to nn do:
m:=i:it0:=0:it1:=0:
for j from 1 to nn while(m<>1) do:
if irem(m,2)=0
then
m:=m/2:it0:=it0+1:
else
m:=3*m+1:it1:=it1+1:
fi:
od:
U[i]:=it0:V[i]:=it1:
od:
for n from 1 to 100 do:
ii:=0:
for k from 1 to nn while(ii=0) do:
if U[k]=U[n]+U[n+1] and V[k]=V[n]+V[n+1]
then
ii:=1:printf(`%d, `,k):
else
fi:
od:
od:
A291352
Numerators of the Collatz parity generator fractions.
Original entry on oeis.org
1, 3, 7, 191, 25983, 2663063519, 7160414743310596607, 208229791652776621349215306687873023871, 45218742589363955614633859184750699518167124046103786511727246869388335095775
Offset: 1
a(1)=1 => f(1) = 1/3 => b(1) = 101010...
a(2)=3 => f(2) = 3/5 => b(2) = 10011001...
elements in b(2) determine the parity of elements at index 2 in the Collatz parity sequence of an integer.
A329535
Numbers with twice as many halving steps before reaching 1 in the 3x + 1 problem as tripling steps.
Original entry on oeis.org
1, 159, 283, 377, 502, 503, 603, 615, 668, 669, 670, 799, 807, 888, 890, 892, 893, 1063, 1065, 1095, 1186, 1187, 1188, 1189, 1190, 1417, 1435, 1580, 1581, 1582, 1585, 1586, 1587, 1889, 1913, 1947, 1959, 1963, 2104, 2106, 2108, 2109, 2113, 2114, 2115, 2119, 2518
Offset: 1
159 is in the sequence because its trajectory, 159, 478, 239, 718, ..., has 36 halving steps and 18 tripling steps.
160 is not in the sequence because its trajectory, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1, has nine even terms but only two odd terms.
-
collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; nn = 50; t = {}; n = 0; While[Length[t] < nn, n++; c = collatz[n]; ev = Length[Select[c, EvenQ]]; od = Length[c] - ev - 1; If[ev == 2 * od, AppendTo[t, n]]]; t
-
def halfTripleCompare(n: Int): Int = {
var curr = n
var htc = 0
while (curr > 1) {
curr = (curr % 2) match {
case 0 => htc = htc + 1
curr / 2
case 1 => htc = htc - 2
3 * curr + 1
}
}
htc
}
(1 to 1000).filter(halfTripleCompare() == 0) // _Alonso del Arte, Nov 18 2019
Comments